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Finding the stationary point of y=lnx/x

I used the quotient rule to differentiate and got (1-lnx)/x^2 which is correct.
The mark scheme then says x=e then y=1/e
How does it get to this step?

Reply 1

Zacken

Original post by ameliamae28

I used the quotient rule to differentiate and got (1-lnx)/x^2 which is correct.
The mark scheme then says x=e then y=1/e
How does it get to this step?

When is (1-ln x)/x^2 = 0? (this is at the stationary point) now multiplying both sides by x^2 gives 1-ln x = 0*x^2 = 0.

So ln x = 1.

So x = e.

Now plug this back into your function y= ln x / x to get the y-value at x=e.

Reply 2

a-spiringmedic

Original post by ameliamae28

I used the quotient rule to differentiate and got (1-lnx)/x^2 which is correct.
The mark scheme then says x=e then y=1/e
How does it get to this step?

after you've differentiated, dy/dx = 0 as at a stationary point the gradient of the curve is 0
can u work it out from there?

Reply 3

the bear

when you have a fraction = zero then the top of the fraction = zero, and you ignore the underneath part...

(3x - 7)/sin(54 -12x) = 0

3x - 7 = 0

x = 7/3